Question: Factor completely. $48-24x+3x^2=$
First, we take a common factor of $3$. $48-24x+3x^2=3(16-8x+x^2)$ Now, let's factor $16-8x+x^2$. Both $16$ and $x^2$ are perfect squares, since $16=({4})^2$ and $x^2=({x})^2$. Additionally, $8x$ is twice the product of the roots of $16$ and $x^2$, since $8x=2({4})( x)$. $16-8x+x^2 = ({4})^2-2({4})( x)+({x})^2$ So we can use the square of a difference pattern to factor: ${a}^2 -2( a)( b)+ {b}^2 =({a}-{b})^2$ In this case, ${a}={4}$ and ${b}={x}$ : $ ({4})^2-2({4})( x)+({x})^2 =({4}-{x})^2$ $\begin{aligned} 48-24x+3x^2&=3(16-8x+x^2) \\\\ &=3(4-x)^2 \end{aligned}$ In conclusion, the complete factorization is $3(4-x)^2$ Remember that you can always check your factorization by expanding it.